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JEE Main 2019 (8 Apr, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2019 (8 Apr, Shift 2)

Q52JEE Main 2019 · 8 APR · SHIFT 2Medium

Electrostatic Potential Energy

A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to :-

Figure

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Answer: (d)

  1. Field of a line charge: E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}.
  2. Work done by the field from r0r_0 to rr: W=∫r0rqλ2πε0rdr=qλ2πε0ln⁡rr0W = \int_{r_0}^{r} \frac{q\lambda}{2\pi\varepsilon_0 r}dr = \frac{q\lambda}{2\pi\varepsilon_0}\ln\frac{r}{r_0}
  3. 12mv2=W⇒v∝ln⁡(rr0)\frac{1}{2}mv^2 = W \Rightarrow v \propto \sqrt{\ln\left(\frac{r}{r_0}\right)}
  4. So the answer is (d).
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